Question

An aerobic, continuous-strirred-tank-reactor system is used to oxidize formaldehyde in a wastewater stream from a particle board manufacturing facility. The initial formaldehyde concentration is 50 mg/L, the degradation rate constant is 0.18/day, the flow rate is 400 gal/min, and the reactor volume is 12,000 gal. Determine (a) the concentration of formaldehyde in the treated effluent and (b) the amount of oxygen required (lb/d) to carry out the process.

Answer 1

for first order reaction in a CSTR ,the characteristic equation is T= CAOXA/-rA= CAO*XA/KCAO*(1-XA)= XA/(K*(1-XA))

KT = XA/(1-XA)

where T= Space time = V/Vo =volume of reactor/ volumetric flow rate= 12000/400= 30 min=30/60 hr= 0.2hr =0.2/24 days=0.00833 days

K= rate constant = 0.18/day

KT= 0.00833*0.18= 0.0015

XA/(1-XA)= 0.0015

XA=0.0015-0.0015XA

1.0015XA= 0.0015

XA= 0.0015/1.0015= 0.001498

XA= Conversion= 1-CA/CAO, CA=Concentration at the outet and CAO= initial concentration=50 mg/L

CA/CAO= 1-0.001498

CA/CAO= 0.9985

CA=0.9985*50=49.925 mg/L

b) Initial concentration of formaldehye= 50 mg/L= 50*10^-3 g/L

flow rate =400 gal/min =400*3.78 5L/min (1 gallon= 3.785 L)= 1514 L/min

mass of Formaldehyde in the feed= 50*10^-3 g/L *1514 L/min =75.7g/min

amount of Fornaldehyde in 24 hrs (1 day)= 75.7*60*24 g/day=109008g/day= 109.008 kg/day

The reaction of Formaldehye (HCHO) and oxygen is given by

HCHO+O2--->CO2 + H2O

Molecular weight of HCHO=1+12+16+1= 30

30 gms requires 32gms of oxygen

109.008 kg/day HCHO requires 109.008*30/32= 116.2752 kg/day of oxygen

but 0.4535 kg= 1 lb

116.2752 kg/day= 116.2752/0.4535= 256.3951 lb/day